Problem: Simplify the following expression: $ x = \dfrac{1}{8} - \dfrac{-2t + 4}{t + 3} $
Solution: In order to subtract expressions, they must have a common denominator. Multiply the first expression by $\dfrac{t + 3}{t + 3}$ $ \dfrac{1}{8} \times \dfrac{t + 3}{t + 3} = \dfrac{t + 3}{8t + 24} $ Multiply the second expression by $\dfrac{8}{8}$ $ \dfrac{-2t + 4}{t + 3} \times \dfrac{8}{8} = \dfrac{-16t + 32}{8t + 24} $ Therefore $ x = \dfrac{t + 3}{8t + 24} - \dfrac{-16t + 32}{8t + 24} $ Now the expressions have the same denominator we can simply subtract the numerators: $x = \dfrac{t + 3 - (-16t + 32) }{8t + 24} $ Distribute the negative sign: $x = \dfrac{t + 3 + 16t - 32}{8t + 24}$ $x = \dfrac{17t - 29}{8t + 24}$